Find the linearization L(x) of f(x) at x = a.
-$$f ( x ) = \sqrt { 4 x + 49 } , a = 0$$
A) $\mathrm { L } ( \mathrm { x } ) = \frac { 2 } { 7 } \mathrm { x } - 7$
B) $L ( x ) = \frac { 4 } { 7 } x + 7$
C) $\mathrm { L } ( \mathrm { x } ) = \frac { 4 } { 7 } \mathrm { x } - 7$
D) $\mathrm { L } ( \mathrm { x } ) = \frac { 2 } { 7 } \mathrm { x } + 7$

Question

Quizplus · Accepted Answer

The linearization of a function $f(x)$ at $x = a$ is given by $L(x) = f(a) + f'(a)(x - a)$. For $f(x) = \sqrt{4x + 49}$, first find $f'(x)$ and then evaluate $f(0)$ and $f'(0)$. $f(0) = \sqrt{49} = 7$ and $f'(x) = \frac{2}{\sqrt{4x + 49}}$, so $f'(0) = \frac{2}{7}$. Thus, $L(x) = 7 + \frac{2}{7}(x - 0) = \frac{2}{7}x + 7$.

Find the Linearization L(x) of F(x) at X = A f(x)=4x+49,a=0f ( x ) = \sqrt { 4 x + 49 } , a = 0f(x)=4x+49​,a=0

Find the Linearization L(x) of F(x) at X = A $f ( x ) = \sqrt { 4 x + 49 } , a = 0$